Integrand size = 27, antiderivative size = 71 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2 c}{27 d^3 \sqrt {c+d x^3}}-\frac {2 \sqrt {c+d x^3}}{3 d^3}+\frac {128 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^3} \] Output:
-2/27*c/d^3/(d*x^3+c)^(1/2)-2/3*(d*x^3+c)^(1/2)/d^3+128/81*c^(1/2)*arctanh (1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^3
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 \left (-\frac {3 \left (10 c+9 d x^3\right )}{\sqrt {c+d x^3}}+64 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )\right )}{81 d^3} \] Input:
Integrate[x^8/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]
Output:
(2*((-3*(10*c + 9*d*x^3))/Sqrt[c + d*x^3] + 64*Sqrt[c]*ArcTanh[Sqrt[c + d* x^3]/(3*Sqrt[c])]))/(81*d^3)
Time = 0.39 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {948, 98, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^6}{\left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx^3\) |
\(\Big \downarrow \) 98 |
\(\displaystyle \frac {1}{3} \int \left (\frac {64 c}{9 d^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}+\frac {c}{9 d^2 \left (d x^3+c\right )^{3/2}}-\frac {1}{d^2 \sqrt {d x^3+c}}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {128 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{27 d^3}-\frac {2 c}{9 d^3 \sqrt {c+d x^3}}-\frac {2 \sqrt {c+d x^3}}{d^3}\right )\) |
Input:
Int[x^8/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]
Output:
((-2*c)/(9*d^3*Sqrt[c + d*x^3]) - (2*Sqrt[c + d*x^3])/d^3 + (128*Sqrt[c]*A rcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(27*d^3))/3
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 1.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(-\frac {2 \left (-64 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \sqrt {d \,x^{3}+c}+27 d \,x^{3}+30 c \right )}{81 \sqrt {d \,x^{3}+c}\, d^{3}}\) | \(54\) |
risch | \(-\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{3}}-\frac {c \left (\frac {2}{27 d \sqrt {d \,x^{3}+c}}-\frac {128 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{81 \sqrt {c}\, d}\right )}{d^{2}}\) | \(60\) |
default | \(-\frac {\frac {2 c}{3 d^{2} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}}{d}+\frac {16 c}{3 d^{3} \sqrt {d \,x^{3}+c}}+\frac {128 c^{2} \left (-\frac {1}{c \sqrt {d \,x^{3}+c}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{27 d^{3}}\) | \(99\) |
elliptic | \(-\frac {2 c}{27 d^{3} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{3}}-\frac {64 i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}-8 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{18 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{243 d^{5}}\) | \(444\) |
Input:
int(x^8/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/81*(-64*c^(1/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))*(d*x^3+c)^(1/2)+27 *d*x^3+30*c)/(d*x^3+c)^(1/2)/d^3
Time = 0.09 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.23 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\left [\frac {2 \, {\left (32 \, {\left (d x^{3} + c\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - 3 \, {\left (9 \, d x^{3} + 10 \, c\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{4} x^{3} + c d^{3}\right )}}, -\frac {2 \, {\left (64 \, {\left (d x^{3} + c\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 3 \, {\left (9 \, d x^{3} + 10 \, c\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{4} x^{3} + c d^{3}\right )}}\right ] \] Input:
integrate(x^8/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
[2/81*(32*(d*x^3 + c)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10* c)/(d*x^3 - 8*c)) - 3*(9*d*x^3 + 10*c)*sqrt(d*x^3 + c))/(d^4*x^3 + c*d^3), -2/81*(64*(d*x^3 + c)*sqrt(-c)*arctan(3*sqrt(-c)/sqrt(d*x^3 + c)) + 3*(9* d*x^3 + 10*c)*sqrt(d*x^3 + c))/(d^4*x^3 + c*d^3)]
Time = 15.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\begin {cases} \frac {2 \left (- \frac {c}{27 \sqrt {c + d x^{3}}} - \frac {64 c \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{81 \sqrt {- c}} - \frac {\sqrt {c + d x^{3}}}{3}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{9}}{72 c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**8/(-d*x**3+8*c)/(d*x**3+c)**(3/2),x)
Output:
Piecewise((2*(-c/(27*sqrt(c + d*x**3)) - 64*c*atan(sqrt(c + d*x**3)/(3*sqr t(-c)))/(81*sqrt(-c)) - sqrt(c + d*x**3)/3)/d**3, Ne(d, 0)), (x**9/(72*c** (5/2)), True))
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left (32 \, \sqrt {c} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 27 \, \sqrt {d x^{3} + c} + \frac {3 \, c}{\sqrt {d x^{3} + c}}\right )}}{81 \, d^{3}} \] Input:
integrate(x^8/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="maxima")
Output:
-2/81*(32*sqrt(c)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*s qrt(c))) + 27*sqrt(d*x^3 + c) + 3*c/sqrt(d*x^3 + c))/d^3
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left (\frac {64 \, c \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d} + \frac {27 \, \sqrt {d x^{3} + c}}{d} + \frac {3 \, c}{\sqrt {d x^{3} + c} d}\right )}}{81 \, d^{2}} \] Input:
integrate(x^8/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
-2/81*(64*c*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d) + 27*sqrt(d* x^3 + c)/d + 3*c/(sqrt(d*x^3 + c)*d))/d^2
Time = 2.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {64\,\sqrt {c}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{81\,d^3}-\frac {2\,c}{27\,d^3\,\sqrt {d\,x^3+c}}-\frac {2\,\sqrt {d\,x^3+c}}{3\,d^3} \] Input:
int(x^8/((c + d*x^3)^(3/2)*(8*c - d*x^3)),x)
Output:
(64*c^(1/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3) ))/(81*d^3) - (2*c)/(27*d^3*(c + d*x^3)^(1/2)) - (2*(c + d*x^3)^(1/2))/(3* d^3)
\[ \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {-\frac {4 \sqrt {d \,x^{3}+c}\, c}{3}-\frac {2 \sqrt {d \,x^{3}+c}\, d \,x^{3}}{3}+8 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c^{2} d^{2}+8 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c \,d^{3} x^{3}}{d^{3} \left (d \,x^{3}+c \right )} \] Input:
int(x^8/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x)
Output:
(2*( - 2*sqrt(c + d*x**3)*c - sqrt(c + d*x**3)*d*x**3 + 12*int((sqrt(c + d *x**3)*x**5)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d**3*x**9),x)*c**2 *d**2 + 12*int((sqrt(c + d*x**3)*x**5)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2 *x**6 - d**3*x**9),x)*c*d**3*x**3))/(3*d**3*(c + d*x**3))